You and your friends are revolving at the same speed. As a result, your friend and you both have the same angular velocity.
Now that we have the formula for linear velocity—V = omega—we can move on to the following section and draw the conclusion that if the relationship between and grows, so will we.
From here, we may also infer that the radius is larger in this instance because you are close to the right edge. Since you and your friend have different linear velocities and you have the higher linear velocity. If you've ever actually been on a merry-go-round, you already know that being on the edge is considerably more enjoyable rotation than being in the centre.
However, there is a similarity between the two people who are moving about. In the same amount of time, they will each cover the same rotation. The term "angular velocity" refers to a speed that measures the angle of rotation over a specified period of time.
To know more about velocity click on the link:
brainly.com/question/18084516
#SPJ4
The cold air increases pressure on the basketball, causing it to appear flat. When it warms up that pressure is taken off, so the basketball is at it's normal state.
Answer:
RT = 17 ohms
Explanation:
For two parallel resistances in a circuit the combined resistance is given by:
R1 = 25 ohms
R2 = 50 ohms
You replace the values of R1 and R2 in the formula for RT:
hence, the combined resitances is 17 ohms
The solution would be like
this for this specific problem:
<span>
The force on m is:</span>
<span>
GMm / x^2 + Gm(2m) / L^2 = 2[Gm (2m) / L^2] ->
1
The force on 2m is:</span>
<span>
GM(2m) / (L - x)^2 + Gm(2m) / L^2 = 2[Gm (2m) / L^2]
-> 2
From (1), you’ll get M = 2mx^2 / L^2 and from
(2) you get M = m(L - x)^2 / L^2
Since the Ms are the same, then
2mx^2 / L^2 = m(L - x)^2 / L^2
2x^2 = (L - x)^2
xsqrt2 = L - x
x(1 + sqrt2) = L
x = L / (sqrt2 + 1) From here, we rationalize.
x = L(sqrt2 - 1) / (sqrt2 + 1)(sqrt2 - 1)
x = L(sqrt2 - 1) / (2 - 1)
x = L(sqrt2 - 1) </span>
= 0.414L
<span>Therefore, the third particle should be located the 0.414L x
axis so that the magnitude of the gravitational force on both particle 1 and
particle 2 doubles.</span>
Answer:
v = 5.26 10² m / s
Explanation:
We can solve this exercise using the concepts of conservation of mechanical energy, because there is no friction
starting point. Red blood cells too far away
Em₀ = K = ½ m v²
final point. Red blood cells touching r = 8.20 10⁻⁶ m
Em_f = U = k q₁ q₂ / r₁₂
Em₀ = Em_f
½ m v² = k q₁ q₂ / r₁₂
v = √ (2 k q₁ q₂ / m r₁₂)
we calculate
v =√ (2 9 10⁹ 2 10⁻¹² 2.9 10⁻¹² / (4.60 10⁻¹⁴ 8.20 10⁻⁶))
v = √ (0.276775 10⁶)
v = 0.526 10³ m / s
v = 5.26 10² m / s
